Is $\{0,1\}^{\omega_1}$ sequentially compact?

It is claimed that an uncountable product of $[0,1]$ is not sequentially compact, e.g. in Wikipedia (I think replacing $[0,1]$ by $\{0,1\}$ doesn't make much difference). However, the constructions I saw always take an uncountable set of the same cardinality as the reals. So I wonder whether $\{0,1\}^{\omega_1}$ is also non-sequentially compact, and if yes, whether one can construct explicitly a sequence without convergent subsequence. Constructing such a sequence in $\{0,1\}^\mathfrak{c}$ requires no choice; does one need some form of choice to say something about $\{0,1\}^{\omega_1}$?

A family $\mathscr{S}\subseteq[\omega]^\omega$ is a splitting family if for each infinite $A\subseteq\omega$ there is an $S\in\mathscr{S}$ such that $A\cap S$ and $A\setminus S$ are both infinite. The splitting number $\mathfrak{s}$ is defined to be the smallest cardinality of a splitting family. It’s easy to prove that $\omega_1\le\mathfrak{s}\le 2^\omega=\mathfrak{c}$. Theorem $6.1$ of Eric K. van Douwen, ‘The Integers and Topology’, in The Handbook of Set-Theoretic Topology, K. Kunen and J.E. Vaughan, eds., North Holland, $1985$, says that
\begin{align*} \mathfrak{s}&=\min\{\kappa:\{0,1\}^\kappa\text{ is not sequentially compact}\}\\ &=\min\{\kappa:\text{there is a compact space of weight }\kappa\text{ that is not sequentially compact}\}\\ &=\min\{\kappa:\text{there is a ctbly cpt space of weight }\kappa\text{ that is not sequentially compact}\}\;. \end{align*}
If $\kappa$ and $\lambda$ are regular cardinals with $\omega_1\le\kappa\le\lambda$ it is consistent with $\mathsf{ZFC}$ that $\mathfrak{s}=\kappa$ and $2^\omega=\lambda$; this is (part of) Theorem $5.1$ of the same monograph.
In short, it $\{0,1\}^{\mathfrak{c}}$ is never sequentially compact, but $\{0,1\}^{\omega_1}$ can be, consistently.
• Also, the Wikipedia page for sequential compactness is misleading: it says "the product of uncountably many copies of the closed unit interval" is not sequentially compact; by the result you quote in your answer (especially the second line), it is consistent with ZFC that $[0,1]^{\omega_1}$ is sequentially compact (the Wiki page cites the book of Steen and Seebach, which in fact discusses $\mathfrak{c}$ many copies of $[0,1]$ - not merely $\omega_1$). This (or an equivalent form thereof) has driven me nuts on and off for quite some time - thank you so much for this answer! – Philip Brooker Nov 7 '13 at 1:19